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typescript - 如何让 TS 在通用构建器函数中推断回调？

问题描述

考虑以下用例演示（游乐场）：

// A builder that can self-reference its keys using a ref function
declare function makeObj<K extends string>(
    builder: (ref: (k: K) => number) => Record<K, number>
): Record<K, number>;

// Not using `ref` for now. All good, K is inferred as <"x" | "y">.
const obj1 = makeObj(() => ({ x: 1, y: 2 }));

// Oops, now that we try to use `ref`, K is inferred as <string>.
const obj2 = makeObj(ref => ({ x: 1, y: ref("invalid key, only x or y") }));

// This works, but we'd want K to be automatically inferred.
const obj3 = makeObj<"x" | "y">(ref => ({ x: 1, y: ref("x") }));

那么，我应该怎么写才能makeObj自动K推断呢？

标签： typescripttypescript-generics

解决方案

尝试声明Record<K, number>为第二种泛型类型。

操场

declare function makeObj<K extends string, T extends Record<K, number> = Record<K, number>>(
  builder: (ref: (k: K) => number) => T
): T

const obj1 = makeObj(() => ({ x: 1, y: 2 }));
const obj2 = makeObj(ref => ({ x: 1, y: ref("invalid key, only x or y") }));
const obj3 = makeObj(ref => ({ x: 1, y: ref("x") }));

局限性

好吧，正如@kaya3 在下面评论的那样。此解析只能推断返回类型。除非显式设置泛型类型，否则它仍然无法找到无效键。

// error will shown when given explicit generic type
const obj2 = makeObj<'x' | 'y'>(ref => ({x: 1, y: ref("invalid key, only x or y")}));

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